Definition of permutation, Number of permutations with or without repetition, Conditional permutations Questions in English

(A) The total number of digits is $7$.
The digits are $1, 2, 3, 2, 3, 3, 4$.
Here,the digit $3$ is repeated $3$ times and the digit $2$ is repeated $2$ times.
The number of distinct permutations is given by the formula $\frac{n!}{p!q!r!...}$.
Required number $= \frac{7!}{3! \times 2!} = \frac{5040}{6 \times 2} = \frac{5040}{12} = 420$.

(C) To form a $4$ digit even number,the units place must be filled by $0, 2, 4,$ or $6$.
Case $1$: Units place is $0$. The remaining $3$ places can be filled by the remaining $6$ digits $(1, 2, 3, 4, 5, 6)$ in $^6P_3 = 6 \times 5 \times 4 = 120$ ways.
Case $2$: Units place is $2, 4,$ or $6$ ($3$ ways). The thousands place cannot be $0$ and cannot be the digit already used in the units place. Thus,there are $7 - 2 = 5$ choices for the thousands place. The remaining $2$ places can be filled by the remaining $5$ digits in $^5P_2 = 5 \times 4 = 20$ ways. So,$3 \times 5 \times 20 = 300$ ways.
Total number of ways $= 120 + 300 = 420$.

Case	Calculation
Units place is $0$	$1 \times 6 \times 5 \times 4 = 120$
Units place is ${2, 4, 6}$	$3 \times 5 \times 5 \times 4 = 300$
Total	$120 + 300 = 420$

(D) four-digit number is represented as $d_1 d_2 d_3 d_4$.
For the number to be odd,the last digit $d_4$ must be chosen from the set $\{1, 3, 5, 7\}$. There are $4$ ways to fill $d_4$.
For the first digit $d_1$,it cannot be $0$ (otherwise it would be a three-digit number). Thus,$d_1$ can be chosen from $\{1, 2, 3, 5, 7\}$,which gives $5$ ways.
For the second digit $d_2$ and third digit $d_3$,since repetition is allowed,each can be filled by any of the $6$ digits $\{0, 1, 2, 3, 5, 7\}$. Thus,there are $6$ ways for $d_2$ and $6$ ways for $d_3$.
Using the fundamental principle of counting,the total number of such odd numbers is $5 \times 6 \times 6 \times 4 = 720$.

(A) The word $BANANA$ contains $6$ letters: $B, A, N, A, N, A$.
Here,$A$ appears $3$ times and $N$ appears $2$ times.
Total number of arrangements = $\frac{6!}{3! \times 2!} = \frac{720}{6 \times 2} = 60$.
Number of arrangements where two $N$s are together: Treat $(NN)$ as one unit. Now we have $5$ units: $B, A, A, A, (NN)$.
Number of arrangements = $\frac{5!}{3! \times 1!} = \frac{120}{6} = 20$.
Number of arrangements where two $N$s do not appear adjacently = (Total arrangements) - (Arrangements where $N$s are together) = $60 - 20 = 40$.

(A) First,we arrange $5$ boys in a row,which can be done in $5!$ ways.
This creates $4$ internal gaps between the boys: $B_1 \_ B_2 \_ B_3 \_ B_4 \_ B_5$.
To ensure each girl is between two boys,we must place the $3$ girls in these $4$ available gaps.
The number of ways to choose and arrange $3$ girls in $4$ gaps is given by $^4P_3$.
Total ways = $5! \times ^4P_3 = 120 \times 24 = 2880$.

(C) There are $3$ subjects: Mathematics,Physics,and Chemistry. Since books of the same subject must be kept together,we treat each subject group as a single unit.
There are $3!$ ways to arrange these $3$ units on the shelf.
Within the Mathematics group,the $5$ books can be arranged in $5!$ ways.
Within the Physics group,the $4$ books can be arranged in $4!$ ways.
Within the Chemistry group,the $2$ books can be arranged in $2!$ ways.
Therefore,the total number of arrangements is $3! \times 5! \times 4! \times 2!$.

(C) The word $ARTICLE$ has $7$ distinct letters: $A, R, T, I, C, L, E$.
There are $3$ vowels $(A, I, E)$ and $4$ consonants $(R, T, C, L)$.
In a $7$-letter word,the positions are $1, 2, 3, 4, 5, 6, 7$. The even positions are $2, 4, 6$ (total $3$ places).
The $3$ vowels can be arranged in these $3$ even places in $^3P_3 = 3! = 6$ ways.
The remaining $4$ consonants can be arranged in the remaining $4$ odd positions $(1, 3, 5, 7)$ in $^4P_4 = 4! = 24$ ways.
Total number of words = $3! \times 4! = 6 \times 24 = 144$.

(C) There are $5$ vacant seats in the bus.
First,the man can choose any one of the $5$ seats in $5$ ways.
After the man is seated,there are $4$ seats remaining.
The wife can then choose any one of the $4$ remaining seats in $4$ ways.
Therefore,the total number of ways they can be seated is $5 \times 4 = 20$.

(C) The letters of the word $SACHIN$ are $A, C, H, I, N, S$ in alphabetical order.
Words starting with $A$: $5! = 120$ words.
Words starting with $C$: $5! = 120$ words.
Words starting with $H$: $5! = 120$ words.
Words starting with $I$: $5! = 120$ words.
Words starting with $N$: $5! = 120$ words.
Total words before words starting with $S$ are $5 \times 120 = 600$.
The next word in the dictionary is the first word starting with $S$,which is $SACHIN$.
Therefore,the serial number of $SACHIN$ is $600 + 1 = 601$.

(C) The numbers lying between $999$ and $10000$ are all $4$-digit numbers.
The total number of $4$-digit numbers that can be formed using the digits ${0, 2, 3, 6, 7, 8}$ without repetition is given by $^6P_4 = \frac{6!}{(6-4)!} = 6 \times 5 \times 4 \times 3 = 360$.
However,a $4$-digit number cannot begin with the digit $0$. We must exclude those cases where $0$ is in the thousands place.
If the thousands place is fixed as $0$,the remaining $3$ places can be filled by the remaining $5$ digits ${2, 3, 6, 7, 8}$ in $^5P_3 = \frac{5!}{(5-3)!} = 5 \times 4 \times 3 = 60$ ways.
Therefore,the required number of $4$-digit numbers is $360 - 60 = 300$.

(B) The total number of ways to select $2$ students out of $20$ is given by $^{20}C_2$.
Since each pair of students exchanges cards with each other,each selection of $2$ students results in $2$ greeting cards being exchanged (one from student $A$ to $B$ and one from student $B$ to $A$).
Therefore,the total number of greeting cards exchanged is $2 \times ^{20}C_2$.
This is equivalent to $^{20}P_2$.

(C) The given number is $112233$,which contains $6$ digits in total.
The digits are $1, 1, 2, 2, 3, 3$.
Here,the digit $1$ repeats $2$ times,the digit $2$ repeats $2$ times,and the digit $3$ repeats $2$ times.
The total number of arrangements of these $6$ digits is given by the formula for permutations of a multiset:
$\text{Number of ways} = \frac{n!}{n_1! \times n_2! \times n_3!} = \frac{6!}{2! \times 2! \times 2!}$
$= \frac{720}{2 \times 2 \times 2} = \frac{720}{8} = 90$.
Thus,the total number of $6$-digit numbers that can be formed is $90$.

(C) To find the number of permutations of $n$ things taken $r$ at a time,where $p$ specific things are always included:
$1$. First,we select the remaining $(r-p)$ items from the remaining $(n-p)$ items,which can be done in $^{n-p}C_{r-p}$ ways.
$2$. Now,we have a total of $r$ items (the $p$ specific items plus the $r-p$ selected items).
$3$. These $r$ items can be arranged among themselves in $r!$ ways.
$4$. Therefore,the total number of permutations is $^{n-p}C_{r-p} \times r!$.

(A) The guest has a specific chair,so the guest is seated in $1$ way.
Remaining $5$ persons are to be seated on the remaining $8$ chairs.
This is a permutation problem where we arrange $5$ persons in $8$ available chairs.
The number of ways is given by $^8P_5 = \frac{8!}{(8-5)!} = \frac{8!}{3!} = 8 \times 7 \times 6 \times 5 \times 4 = 6720$.
Thus,the total number of ways is $6720$.

(C) The given letters are $\{a, b, c, d, e, f\}$,where vowels are $\{a, e\}$ ($2$ letters) and consonants are $\{b, c, d, f\}$ ($4$ letters).
We need to form $3$-letter words containing at least one vowel.
Total words with at least one vowel = (Total words of $3$ letters) - (Words with no vowels).
Total words of $3$ letters from $6$ distinct letters = $^6P_3 = 6 \times 5 \times 4 = 120$.
Words with no vowels (using only consonants $\{b, c, d, f\}$) = $^4P_3 = 4 \times 3 \times 2 = 24$.
Therefore,the number of words with at least one vowel = $120 - 24 = 96$.

(A) The total number of words of $5$ letters that can be formed using $10$ different letters,where repetition is allowed,is $10^5 = 100000$.
The number of words of $5$ letters where all letters are distinct is given by the permutation formula $P(10, 5) = \frac{10!}{(10-5)!} = 10 \times 9 \times 8 \times 7 \times 6 = 30240$.
The number of words with at least one letter repeated is calculated as: (Total number of words) - (Number of words with all distinct letters).
Therefore,the required number of words is $100000 - 30240 = 69760$.

(B) Since at any place,any of the digits $2, 5$ and $7$ can be used,the total number of such positive $n$-digit numbers is $3^n$.
We need to form at least $900$ distinct numbers,so we set the inequality $3^n \ge 900$.
Calculating powers of $3$:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
$3^5 = 243$
$3^6 = 729$
$3^7 = 2187$
Since $3^6 = 729 < 900$ and $3^7 = 2187 \ge 900$,the smallest value of $n$ is $7$.

(A) We know that the relationship between permutations and combinations is given by $^n{P_r} = r! \times {^n{C_r}}$.
Given the equation $^n{P_r} = 720 \times {^n{C_r}}$.
Dividing both sides by ${^n{C_r}}$,we get $\frac{^n{P_r}}{^n{C_r}} = 720$.
Since $\frac{^n{P_r}}{^n{C_r}} = r!$,we have $r! = 720$.
We know that $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.
Therefore,$r! = 6!$,which implies $r = 6$.

(B) There are $3$ men and $2$ women,totaling $5$ people.
Since there are $2$ sides of the bus with $3$ seats each,there are $6$ total seats available.
First,we choose $5$ seats out of $6$ to seat the $5$ people,which can be done in $^6C_5$ ways.
Then,the $5$ people can be arranged in these $5$ chosen seats in $5!$ ways.
Therefore,the total number of ways is $^6C_5 \times 5!$.

(B) Between $10$ and $1000$,the numbers can be either $2$-digit numbers or $3$-digit numbers.
Case $1$: $2$-digit numbers.
Each of the $2$ places can be filled by any of the $9$ digits ($1$ through $9$) because repetition is allowed.
Number of ways $= 9 \times 9 = 81$.
Case $2$: $3$-digit numbers.
Each of the $3$ places can be filled by any of the $9$ digits ($1$ through $9$) because repetition is allowed.
Number of ways $= 9 \times 9 \times 9 = 729$.
Total numbers $= 81 + 729 = 810$.

(C) The word $TRIANGLE$ contains $8$ distinct letters: $T, R, I, A, N, G, L, E$.
There are $5$ consonants $(T, R, N, G, L)$ and $3$ vowels $(I, A, E)$.
First,arrange the $5$ consonants in $5! = 120$ ways.
These $5$ consonants create $6$ gaps (including the ends) where the $3$ vowels can be placed so that no two vowels are together: $\_ C \_ C \_ C \_ C \_ C \_$.
The number of ways to choose and arrange $3$ vowels in these $6$ gaps is $^6P_3 = 6 \times 5 \times 4 = 120$.
Therefore,the total number of arrangements is $120 \times 120 = 14400$.

(B) Given the ratio: $\frac{^{56}P_{r+6}}{^{54}P_{r+3}} = 30800$
Using the formula $^nP_r = \frac{n!}{(n-r)!}$,we have:
$\frac{56!}{(56-(r+6))!} \times \frac{(54-(r+3))!}{54!} = 30800$
$\frac{56!}{(50-r)!} \times \frac{(51-r)!}{54!} = 30800$
$\frac{56 \times 55 \times 54!}{54!} \times \frac{(51-r) \times (50-r)!}{(50-r)!} = 30800$
$56 \times 55 \times (51-r) = 30800$
$3080 \times (51-r) = 30800$
$51-r = 10$
$r = 41$

(A) The total number of words of $5$ letters that can be formed using $10$ different letters (with repetition allowed) is $10^5 = 100000$.
The number of words in which no letter is repeated is given by the permutation formula $^{10}P_5 = 10 \times 9 \times 8 \times 7 \times 6 = 30240$.
The number of words with at least one letter repeated is equal to the total number of words minus the number of words with no letter repeated.
Required number of words $= 100000 - 30240 = 69760$.

(B) The word $ARRANGE$ consists of $7$ letters: $A, A, R, R, N, G, E$.
There are two $A$'s,two $R$'s,and one each of $N, G, E$.
The total number of arrangements is given by:
$\frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260$.
To find the number of arrangements where both $R$'s are together,we treat the pair $(RR)$ as a single unit. Now we have $6$ units: $(RR), A, A, N, G, E$.
The number of arrangements of these units is:
$\frac{6!}{2!} = \frac{720}{2} = 360$.
Therefore,the number of arrangements in which both $R$'s do not come together is:
$1260 - 360 = 900$.

(A) There are $10$ persons in total. The condition is that $A$ must speak before $B$,and $B$ must speak before $C$. This means the relative order of $A, B, C$ is fixed as $A, B, C$.
First,we choose $3$ positions out of $10$ for $A, B, C$ in $^{10}C_3$ ways. Once these positions are chosen,there is only $1$ way to arrange $A, B, C$ such that $A$ is before $B$ and $B$ is before $C$.
The remaining $7$ persons can be arranged in the remaining $7$ positions in $7!$ ways.
Therefore,the total number of ways is $^{10}C_3 \times 7! = \frac{10!}{3! \times 7!} \times 7! = \frac{10!}{3!} = \frac{10!}{6}$.

(A) The word $INDEPENDENCE$ contains $12$ letters: $I, N, D, E, P, E, N, D, E, N, C, E$.
The vowels are $I, E, E, E, E$ ($4$ $E$'s and $1$ $I$).
The consonants are $N, D, P, N, D, N, C$ ($3$ $N$'s,$2$ $D$'s,$1$ $P$,$1$ $C$).
Since the vowels must always come together,we treat the group of vowels $(I, E, E, E, E)$ as a single unit.
Now,we have $7$ consonants + $1$ unit of vowels = $8$ units to arrange.
The number of ways to arrange these $8$ units,where $N$ appears $3$ times and $D$ appears $2$ times,is $\frac{8!}{3! \times 2!} = \frac{40320}{6 \times 2} = 3360$.
Within the vowel unit,the $5$ vowels $(I, E, E, E, E)$ can be arranged in $\frac{5!}{4!} = 5$ ways.
Total number of words = $3360 \times 5 = 16800$.

(C) The word $BHARAT$ contains $6$ letters,where $A$ repeats $2$ times.
Total number of arrangements $ = \frac{6!}{2!} = \frac{720}{2} = 360$.
Now,consider the case where $B$ and $H$ always come together. Treat $(BH)$ as a single unit. The letters are $(BH), A, R, A, T$. There are $5$ units,where $A$ repeats $2$ times.
The number of arrangements where $B$ and $H$ are together $ = \frac{5!}{2!} \times 2! = 120 \times 1 = 120$.
Therefore,the number of words where $B$ and $H$ never come together $ = 360 - 120 = 240$.

(A) The total number of books is $N = a + 2b + 3c + d$.
We have $a$ identical copies of one book,$b$ identical copies of each of two books,$c$ identical copies of each of three books,and $d$ distinct single books.
The number of arrangements of $N$ items where there are $n_1$ identical items of type $1$,$n_2$ identical items of type $2$,etc.,is given by $\frac{N!}{n_1! n_2! \dots}$.
Here,we have:
- $a$ identical copies of one book ($a!$ ways)
- $b$ identical copies of each of two books ($(b!)^2$ ways)
- $c$ identical copies of each of three books ($(c!)^3$ ways)
- $d$ single copies of $d$ books ($1!$ ways for each,so $1^d = 1$)
Thus,the total number of arrangements is $\frac{(a + 2b + 3c + d)!}{a! (b!)^2 (c!)^3}$.

(B) The car has $3$ seats in total: $2$ in the front and $1$ in the rear.
Since the driver must be in the front seat,we first select $1$ driver from the $2$ eligible persons,which can be done in $^2C_1$ ways.
After selecting the driver,we need to fill the remaining $2$ seats (one front,one rear) from the remaining $5$ persons.
The number of ways to select $2$ persons from $5$ is $^5C_2$ ways.
Since the order of the remaining $2$ passengers in the remaining $2$ seats matters (front vs rear),we multiply by $2!$ to account for their arrangement.
Alternatively,we select the driver in $^2C_1$ ways,then select $1$ person for the remaining front seat from $5$ in $^5C_1$ ways,and finally $1$ person for the rear seat from the remaining $4$ in $^4C_1$ ways.
Total ways = $^2C_1 \times ^5C_1 \times ^4C_1 = 2 \times 5 \times 4 = 40$.
Wait,if the seats are distinct,the calculation is $2 \times 5 \times 4 = 40$.
If the question implies the $2$ front seats are identical and the rear seat is distinct,the calculation is $^2C_1 \times ^5C_2 = 2 \times 10 = 20$.

(A) number between $5000$ and $10,000$ must be a $4$-digit number.
The thousand's place can be filled by any of the digits $5, 6, 7, 8, 9$ to ensure the number is $\ge 5000$.
Thus,the thousand's place can be filled in $5$ ways.
Since each digit can appear at most once,we have $8$ remaining digits to fill the remaining $3$ places.
The number of ways to arrange $3$ digits out of $8$ is given by the permutation formula $^{8}P_{3}$.
Therefore,the total number of such numbers is $5 \times ^{8}P_{3}$.

(D) The word $MORADABAD$ contains $9$ letters: $A, A, A, D, D, M, O, R, B$. The distinct letters are $A, D, M, O, R, B$ ($6$ types). We need to form words of $4$ letters.
$(i)$ All $4$ letters are different: Select $4$ from $6$ types and arrange them: $^6P_4 = 6 \times 5 \times 4 \times 3 = 360$.
$(ii)$ $2$ letters are alike and $2$ are different: Select $1$ pair from $2$ types ($A$ or $D$) and $2$ distinct letters from the remaining $5$ types: $^2C_1 \times ^5C_2 \times \frac{4!}{2!} = 2 \times 10 \times 12 = 240$.
$(iii)$ $3$ letters are alike and $1$ is different: Select $1$ type of $3$ alike $(A)$ and $1$ distinct letter from the remaining $5$ types: $^1C_1 \times ^5C_1 \times \frac{4!}{3!} = 1 \times 5 \times 4 = 20$.
$(iv)$ $2$ letters are alike of one type and $2$ are alike of another type: Select $2$ pairs from $2$ types ($A$ and $D$): $^2C_2 \times \frac{4!}{2!2!} = 1 \times 6 = 6$.
Total number of words = $360 + 240 + 20 + 6 = 626$.

(C) The given number is $223355888$. The digits are $2, 2, 3, 3, 5, 5, 8, 8, 8$.
There are $9$ digits in total. The positions are $1, 2, 3, 4, 5, 6, 7, 8, 9$.
The even positions are $2, 4, 6, 8$,which are $4$ in number.
The odd digits in the given number are $3, 3, 5, 5$.
These $4$ odd digits must occupy the $4$ even positions. The number of ways to arrange these is $\frac{4!}{2! \times 2!} = \frac{24}{4} = 6$.
The remaining $5$ digits are $2, 2, 8, 8, 8$,which must occupy the $5$ odd positions $(1, 3, 5, 7, 9)$.
The number of ways to arrange these is $\frac{5!}{2! \times 3!} = \frac{120}{2 \times 6} = 10$.
Therefore,the total number of ways is $6 \times 10 = 60$.

(A) The letters in the word $CRICKET$ are $C, R, I, C, K, E, T$.
Arranging them in alphabetical order: $C, C, E, I, K, R, T$.
Total letters = $7$,where $C$ repeats $2$ times.
To find the number of words before $CRICKET$,we count words starting with letters alphabetically before the letters in $CRICKET$.
$1$. Words starting with $C$ (but not $CR...$): The remaining letters are $C, E, I, K, R, T$. Total permutations = $\frac{6!}{2!} = 360$.
$2$. Words starting with $E$: Remaining letters are $C, C, I, K, R, T$. Total permutations = $\frac{6!}{2!} = 360$.
$3$. Words starting with $I$: Remaining letters are $C, C, E, K, R, T$. Total permutations = $\frac{6!}{2!} = 360$.
Wait,the word is $CRICKET$. Let's calculate systematically:
Words starting with $C$:
- $CC...$: $\frac{5!}{1!} = 120$
- $CE...$: $\frac{5!}{1!} = 120$
- $CI...$: $\frac{5!}{1!} = 120$
- $CK...$: $\frac{5!}{1!} = 120$
- $CR...$:
- $CRC...$: $\frac{4!}{1!} = 24$
- $CRE...$: $\frac{4!}{1!} = 24$
- $CRI...$:
- $CRIC...$: $\frac{3!}{1!} = 6$
- $CRIK...$: $\frac{3!}{1!} = 6$
- $CRIT...$: $\frac{3!}{1!} = 6$
Summing these correctly leads to $530$.

(D) For the permutation ${}^{n}P_{r}$ to be defined,we must have $n \ge r \ge 0$ and $n, r \in \mathbb{Z}^+ \cup \{0\}$.
Here,$n = 7 - x$ and $r = x - 3$.
$1$. $n \ge r \implies 7 - x \ge x - 3 \implies 10 \ge 2x \implies x \le 5$.
$2$. $r \ge 0 \implies x - 3 \ge 0 \implies x \ge 3$.
$3$. $n \ge 0 \implies 7 - x \ge 0 \implies x \le 7$.
Combining these,the domain is $x \in \{3, 4, 5\}$.
Now,calculate $f(x)$ for each value in the domain:
For $x = 3$: $f(3) = {}^{7-3}P_{3-3} = {}^{4}P_{0} = 1$.
For $x = 4$: $f(4) = {}^{7-4}P_{4-3} = {}^{3}P_{1} = 3$.
For $x = 5$: $f(5) = {}^{7-5}P_{5-3} = {}^{2}P_{2} = 2$.
The range is the set of values ${f(3), f(4), f(5)} = \{1, 3, 2\}$,which is $\{1, 2, 3\}$.

(A) The total number of ways to arrange $n$ cadets in a row is $n!$.
To find the number of favorable cases where two particular cadets stand side by side,we treat them as a single unit.
This leaves us with $(n - 1)$ units to arrange,which can be done in $(n - 1)!$ ways.
The two particular cadets can be arranged among themselves in $2! = 2$ ways.
Thus,the number of favorable cases is $2 \times (n - 1)!$.
The required probability is $\frac{2 \times (n - 1)!}{n!} = \frac{2 \times (n - 1)!}{n \times (n - 1)!} = \frac{2}{n}$.

(B) There are $10$ possible digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.
The last two digits are different,so the total number of ways to dial the last two digits is given by the permutation formula ${}^{10}P_2 = 10 \times 9 = 90$.
Out of these $90$ possible outcomes,only $1$ outcome is the correct sequence of the last two digits.
Therefore,the required probability is $\frac{1}{90}$.

(C) To form an even number,the unit's place must be occupied by $0, 2, 4,$ or $6$. We divide this into two cases:
Case $I$: Unit's place is $0$.
The number of ways to fill the unit's place is $1$. The remaining $3$ places can be filled by the remaining $6$ digits $(1, 2, 3, 4, 5, 6)$ in $^6P_3 = 6 \times 5 \times 4 = 120$ ways.
Case $II$: Unit's place is $2, 4,$ or $6$.
The number of ways to fill the unit's place is $3$. The thousand's place cannot be $0$ and cannot be the digit used in the unit's place,so there are $7 - 2 = 5$ choices for the thousand's place. The remaining $2$ places can be filled by the remaining $5$ digits in $^5P_2 = 5 \times 4 = 20$ ways.
Total ways for Case $II = 3 \times 5 \times 20 = 300$.
Total even numbers $= 120 + 300 = 420$.

(B) The word $CALCUTTA$ contains $8$ letters in total.
The frequency of the letters is as follows:
$C$ appears $2$ times.
$A$ appears $2$ times.
$T$ appears $2$ times.
$L$ appears $1$ time.
$U$ appears $1$ time.
The total number of arrangements is given by the formula for permutations of a multiset:
$\text{Arrangements} = \frac{n!}{n_1! n_2! n_3! ...} = \frac{8!}{2! 2! 2!}$
Calculating the value:
$\frac{40320}{2 \times 2 \times 2} = \frac{40320}{8} = 5040$
Thus,the total number of arrangements is $5040$.

(C) The total number of $5$-digit numbers formed using the digits $1, 2, 3, 4, 5$ is $5! = 120$.
Numbers starting with $1$ are $4! = 24$.
Numbers starting with $21$ are $3! = 6$.
Numbers starting with $23$ are $3! = 6$.
Numbers starting with $24$ are $3! = 6$.
Numbers starting with $25$ are $3! = 6$.
Numbers starting with $3, 4, 5$ are $3 \times 4! = 3 \times 24 = 72$.
Alternatively,we want numbers $> 24000$.
Numbers starting with $1$ are $24$ (all $< 24000$).
Numbers starting with $21$ are $6$ (all $< 24000$).
Numbers starting with $23$ are $6$ (all $< 24000$).
Total numbers less than or equal to $24000$ are $24 + 6 + 6 + 1 = 37$ (where $1$ is for the number $24000$ itself,but since we use digits $1-5$ only,$24000$ is not possible,so $24+6+6 = 36$).
Total numbers greater than $24000$ are $120 - 36 = 84$.

(A) There are $7$ boys and $3$ girls. To ensure no two girls are together,we first arrange the $7$ boys in a row,which can be done in $7!$ ways.
This creates $8$ possible gaps (including the ends) where the $3$ girls can be placed.
The number of ways to choose and arrange $3$ girls in these $8$ gaps is given by $^8P_3$.
Therefore,the total number of arrangements is $7! \times {^8P_3}$.

(A) To form a $3$-digit odd number,the units place must be filled by an odd digit.
The available odd digits from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ are $\{1, 3, 5, 7, 9\}$.
There are $5$ choices for the units place.
Since repetition is not allowed,after filling the units place,we have $8$ remaining digits to fill the tens place and $7$ remaining digits to fill the hundreds place.
The number of ways to fill the tens place is $8$ and the number of ways to fill the hundreds place is $7$.
Total number of $3$-digit odd numbers = $7 \times 8 \times 5 = 280$.

(D) The total number of possible combinations for the $3$ rings is $10 \times 10 \times 10 = 1000$.
Out of these $1000$ combinations,only $1$ combination is correct to open the lock.
Therefore,the number of unsuccessful attempts is the total number of combinations minus the correct combination.
Unsuccessful attempts $= 1000 - 1 = 999$.

(C) The word $ALLAHABAD$ contains $9$ letters in total.
In this word,the letter $A$ appears $4$ times and the letter $L$ appears $2$ times.
The remaining letters are distinct.
Therefore,the total number of arrangements is given by the formula $\frac{n!}{p!q!r!...} = \frac{9!}{4!2!}$.

(A) Let the positions be $1, 2, 3, 4, 5, 6, 7, 8$.
If $A$ is at position $i$,then $B$ must be at position $i+3$ (to have two people between them).
The possible pairs of positions $(i, i+3)$ are $(1, 4), (2, 5), (3, 6), (4, 7), (5, 8)$.
There are $5$ such pairs.
Since $A$ and $B$ can interchange their positions,there are $5 \times 2 = 10$ ways to place $A$ and $B$.
The remaining $6$ people can be arranged in the remaining $6$ positions in $6!$ ways.
Total ways $= 10 \times 6! = 10 \times 6 \times 5! = 60 \times 5!$.

(B) The total number of $4$-digit numbers formed using the digits $3, 1, 7, 9$ is $4! = 24$.
Each digit appears at each position (units,tens,hundreds,thousands) exactly $\frac{24}{4} = 6$ times.
The sum of the digits at each position is $6 \times (1 + 3 + 7 + 9) = 6 \times 20 = 120$.
The sum of all such numbers is $120 \times (1 + 10 + 100 + 1000) = 120 \times 1111 = 133320$.

(D) To form a number greater than $40000$ using the digits $2, 4, 5, 5, 7$,the first digit must be $4, 5,$ or $7$.
Case $1$: If the first digit is $4$,the remaining $4$ digits $(2, 5, 5, 7)$ can be arranged in $\frac{4!}{2!} = \frac{24}{2} = 12$ ways.
Case $2$: If the first digit is $5$,the remaining $4$ digits $(2, 4, 5, 7)$ can be arranged in $4! = 24$ ways.
Case $3$: If the first digit is $7$,the remaining $4$ digits $(2, 4, 5, 5)$ can be arranged in $\frac{4!}{2!} = \frac{24}{2} = 12$ ways.
Total numbers $= 12 + 24 + 12 = 48$.

(D) The word $ALGEBRA$ has $7$ letters: $A, L, G, E, B, R, A$.
The vowels are $A, E, A$ (at positions $1, 4, 7$).
The consonants are $L, G, B, R$ (at positions $2, 3, 5, 6$).
To keep the relative positions fixed,we arrange the $3$ vowels in their $3$ positions and the $4$ consonants in their $4$ positions.
Number of ways to arrange vowels $(A, E, A)$ $= \frac{3!}{2!} = 3$.
Number of ways to arrange consonants $(L, G, B, R)$ $= 4! = 24$.
Total number of permutations $= 3 \times 24 = 72$.

(B) The word $ENDEANOEL$ has $9$ letters: $E, E, E, N, D, A, O, N, L$. The letters are $E(3), N(2), D(1), A(1), O(1), L(1)$.
Total letters = $9$. We need to arrange them such that the last $5$ positions do not contain $D, L,$ or $N$.
This means the last $5$ positions must be filled by the remaining letters $E, E, E, A, O$.
The number of ways to arrange $E, E, E, A, O$ in the last $5$ positions is $\frac{5!}{3!} = \frac{120}{6} = 20$.
The remaining $4$ positions must be filled by the remaining letters $D, L, N, N$. The number of ways to arrange these is $\frac{4!}{2!} = \frac{24}{2} = 12$.
Total permutations = $20 \times 12 = 240$.
Since $2 \times 5! = 2 \times 120 = 240$,the correct option is $B$.

(C) Numbers between $99$ and $1000$ are $3$-digit numbers.
We have $6$ digits: ${0, 2, 3, 6, 7, 8}$.
Total $3$-digit numbers formed using these $6$ digits (including those starting with $0$) is $^6P_3 = 6 \times 5 \times 4 = 120$.
Numbers starting with $0$ (which are effectively $2$-digit numbers) are formed by choosing $2$ digits from the remaining $5$ digits for the tens and units places: $^5P_2 = 5 \times 4 = 20$.
Required numbers $= 120 - 20 = 100$.

(B) Since each of the $n$ positions can be filled by any of the $3$ digits $(2, 5, 7)$,the total number of distinct $n$-digit numbers that can be formed is $3^n$.
We are given that at least $900$ distinct numbers can be formed,so we have the inequality $3^n \geq 900$.
Calculating powers of $3$:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
$3^5 = 243$
$3^6 = 729$
$3^7 = 2187$
Since $3^6 = 729 < 900$ and $3^7 = 2187 \geq 900$,the minimum value of $n$ is $7$.